Bourbaki

Exercise 5, § 3

General Topology, IX

Bourbaki marks this one as harder than the rest, it is set in small type as supplementary.

Let ω\omega be an absolute value on the field Q\mathbf{Q} of rational numbers. a) Show that ω\omega is uniquely determined by its values at the prime numbers. b) If there is a prime number pp such that ω(p)≤1\omega(p) \leq 1, show that ω(q)≤1\omega(q) \leq 1 for every other prime number qq [find an upper bound for ω(qn)\omega(q^n) by writing qnq^n in the scale of pp, and let n→∞n \to \infty]. c) If there exists a prime number pp such that ω(p)<1\omega(p) < 1, show that ω(q)=1\omega(q) = 1 for every other prime number qq [show that we cannot have ω(q)<1\omega(q) < 1, by using the fact that for each integer n>0n > 0 there exist two rational integers rr and ss such that 1=rpn+sqn1 = rp^n + sq^n]. Hence show that ω\omega is then an absolute value equivalent to the pp-adic absolute value on Q\mathbf{Q}. d) If ω(p)>1\omega(p) > 1 for every prime number pp, show that if pp and qq are any two prime numbers, then

log⁡ω(p)log⁡p=log⁡ω(q)log⁡q \frac{\log \omega(p)}{\log p} = \frac{\log \omega(q)}{\log q}
[same method as in b)]. Hence show that in this case ω(x)=∣x∣p\omega(x) = |x|^p for some p≤1p \leq 1. \*

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.